Magnetisation and Magnetic Intensity

Magnetisation: How Strongly Is a Sample Magnetised?

The world is full of elements, compounds and alloys — and we'd like to classify how each responds to a magnetic field. This section builds the vocabulary for that.

Recall that a circulating electron in an atom has a magnetic moment. In a bulk material these atomic moments add vectorially, and the sample can end up with a net moment. We define the magnetisation $\vec{M}$ of a sample as its net magnetic moment per unit volume:

$$M = \frac{m_{net}}{V}$$

• $\vec{M}$ is a vector.
• Dimensions: L$^{-1}$A; SI unit: A m$^{-1}$ (ampere per metre).

Quick feel for it: a 1 cm$^3$ sample ($10^{-6}$ m$^3$) with net moment $10^{-3}$ A m$^2$ has $M = 10^3$ A/m.

A Solenoid With a Core

Consider a long solenoid, $n$ turns per unit length, current $I$. From Chapter 4, the interior field (vacuum inside) is

$$B_0 = \mu_0 n I$$

Now fill the interior with a material of non-zero magnetisation. The field inside changes because the magnetised material contributes its own magnetic field:

$$\vec{B} = \vec{B}_0 + \vec{B}_m$$

For the long-solenoid geometry used here, the material contribution is

$$\vec{B}_m = \mu_0 \vec{M}$$

where $\mu_0$ is the same vacuum permeability that appears in the Biot-Savart law. If $\vec{M}$ is along the solenoid field, $B$ becomes larger than $B_0$; if $\vec{M}$ is opposite to it, as in diamagnetic response, $B$ becomes slightly smaller.

So the total field has two distinct parents: the external current (giving $\vec{B}_0$) and the material's own response (giving $\vec{B}_m$). Keeping these separate is exactly what the next quantity is for.

Magnetic Intensity H: Separating Cause From Response

Define the magnetic intensity $\vec{H}$ by

$$\vec{H} = \frac{\vec{B}}{\mu_0} - \vec{M}$$

so that the total field can always be written as

$$\vec{B} = \mu_0(\vec{H} + \vec{M})$$

For a long solenoid where all vectors are along the axis, this is often written in scalar form as $B = \mu_0(H+M)$, with signs/directions understood.

The partition is beautifully clean:

• $\vec{H}$ represents the part due to external factors — the free current in the windings. For our solenoid, $H = nI$, whether or not a core is present.
• $\vec{M}$ represents the part due to the specific nature of the magnetic material.
• $\vec{H}$ has the same dimensions as $\vec{M}$: unit A m$^{-1}$.

Key Point: Same coil, same current — same $H$. Insert a core and $B$ may grow hundreds of times, but $H$ doesn't budge. $H$ is the cause; $M$ is the material's response; $B$ is the total effect.

[NEET Important] Be careful with the trio of names: $B$ = magnetic field (tesla), $H$ = magnetic intensity (A/m), $M$ = magnetisation (A/m). Mixing up their units is the classic trap.

Susceptibility, Relative Permeability, Permeability

How strongly does the material respond to a given $H$? For most linear magnetic materials, especially diamagnetic and paramagnetic materials in ordinary fields, the response is written as

$$\vec{M} = \chi \vec{H}$$

where $\chi$ is the magnetic susceptibility — dimensionless, a pure number measuring how a material responds to an external field.

• $\chi$ small and positive (about $+10^{-5}$): paramagnetic ($\vec{M}$ along $\vec{H}$).
• $\chi$ small and negative (about $-10^{-5}$): diamagnetic ($\vec{M}$ opposite to $\vec{H}$).

Substituting into $B = \mu_0(H+M)$ for a linear isotropic material:

$$B = \mu_0(1+\chi)H = \mu_0\mu_r H = \mu H$$

where

$$\mu_r = 1 + \chi \qquad \text{and} \qquad \mu = \mu_0\mu_r$$

• $\mu_r$ = relative magnetic permeability: dimensionless, the magnetic analog of the dielectric constant in electrostatics.
• $\mu$ = magnetic permeability: same dimensions and units as $\mu_0$ (T m A$^{-1}$).

[JEE Tip] $\chi$, $\mu_r$ and $\mu$ are interrelated — only one is independent. Given any one, write the other two instantly: $\mu_r = 1+\chi$, $\mu = \mu_0(1+\chi)$. Exam questions love disguising this one-liner as a hard problem.

Solved Examples

Example 1: H inside a cored solenoid

A solenoid has 1000 turns per metre and carries a current of 2.0 A. Its core has relative permeability 400. Find the magnetic intensity H inside.

Solution:

1. Key fact: $H$ depends only on the free current, not on the core: $H = nI$.
2. $H = 1000 \times 2.0 = 2 \times 10^3$ A/m.
3. Answer: $H = 2 \times 10^3$ A/m — identical with or without the core.

Example 2: The field B inside

For the same solenoid, find the magnetic field B inside the core.

Solution:

1. Formula: $B = \mu_r \mu_0 H$.
2. $B = 400 \times (4\pi \times 10^{-7}) \times 2 \times 10^3$.
3. $4\pi \times 10^{-7} \times 2 \times 10^3 = 2.5 \times 10^{-3}$; times 400 gives about 1.0.
4. Answer: $B \approx 1.0$ T — the core boosts the field 400-fold.

Example 3: Magnetisation of the core

Find the magnetisation M of the core in the previous example.

Solution:

1. From the master relation: $M = \frac{B - \mu_0 H}{\mu_0} = (\mu_r - 1)H$.
2. $M = 399 \times 2 \times 10^3 \approx 8 \times 10^5$ A/m.
3. Check the hierarchy: $M \gg H$ here — in a strongly magnetic core, almost all of B comes from the material itself.

Example 4: The magnetising current

What additional current $I_m$, passed through the same windings without the core, would produce the same B?

Solution:

1. Set up: $B = \mu_0 n (I + I_m)$ with $B = 1.0$ T.
2. $I + I_m = \frac{B}{\mu_0 n} = \frac{1.0}{4\pi \times 10^{-7} \times 1000} \approx 796$ A.
3. With $I = 2$ A: $I_m \approx 794$ A.
4. Takeaway: the core does the work of an extra 794 A! That's why electromagnets use iron cores instead of monstrous currents.

Example 5: From susceptibility to the rest

A material has magnetic susceptibility $\chi = 5.9 \times 10^{-4}$. Find its relative permeability and permeability, and classify it.

Solution:

1. $\mu_r = 1 + \chi = 1.00059$.
2. $\mu = \mu_0 \mu_r = 4\pi \times 10^{-7} \times 1.00059 \approx 1.257 \times 10^{-6}$ T m/A.
3. $\chi$ is small and positive: the material is paramagnetic.

Example 6: Magnetisation from moment and volume

A sample of volume $10^{-5}$ m$^3$ has a net magnetic moment of $8 \times 10^{-2}$ A m$^2$. Find its magnetisation.

Solution:

1. Formula: $M = m_{net}/V$.
2. $M = \frac{8 \times 10^{-2}}{10^{-5}} = 8 \times 10^3$ A/m.
3. Answer: $M = 8 \times 10^3$ A/m along the direction of the net moment.

Example 7: Paramagnetic response numerical

Aluminium has $\chi \approx 2.3 \times 10^{-5}$. A field intensity $H = 10^4$ A/m is applied. Find M and B inside the sample.

Solution:

1. $M = \chi H = 2.3 \times 10^{-5} \times 10^4 = 0.23$ A/m — tiny but along H.
2. $B = \mu_0(H + M) \approx \mu_0 H (1+\chi) \approx 4\pi \times 10^{-7} \times 10^4 = 1.26 \times 10^{-2}$ T.
3. The correction from the material is just 2.3 parts in $10^5$ — paramagnetism is weak.

Example 8: Diamagnetic response

For copper, $\chi \approx -9.8 \times 10^{-6}$. With the same $H = 10^4$ A/m, what are M and B?

Solution:

1. $M = \chi H = -9.8 \times 10^{-6} \times 10^4 = -0.098$ A/m — opposite to H (the meaning of negative $\chi$).
2. $B = \mu_0(1+\chi)H$ is slightly less than $\mu_0 H$: reduced by about 1 part in $10^5$.
3. Takeaway: in diamagnets, M opposes H, so the material slightly expels the field.

Example 9: Finding H from B in a core

The field inside a material of relative permeability 1000 is 0.5 T. Find H and M.

Solution:

1. $H = \frac{B}{\mu_0\mu_r} = \frac{0.5}{4\pi \times 10^{-7} \times 1000} \approx 398$ A/m.
2. $M = (\mu_r - 1)H = 999 \times 398 \approx 3.98 \times 10^5$ A/m.
3. Sanity check: $\mu_0(H+M) = 4\pi \times 10^{-7} \times (398 + 3.98 \times 10^5) \approx 0.5$ T. Consistent.

Example 10: Why bother with H at all?

A student asks: 'B already describes the field. Why invent H?'

Solution:

1. $B$ inside matter mixes two contributions — the coil's current and the material's response — which change together when you swap cores.
2. $H$ isolates what the experimenter controls: for a solenoid $H = nI$, fixed by the winding and current alone.
3. Given $H$ and the material's $\chi$, everything follows: $M = \chi H$, $B = \mu_0(1+\chi)H$. One controlled input, one material property, total field predicted.